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Inequalities from 2008 Mathematical Competition Inequalities from 2008 Mathematical Competition Editor Manh Dung Nguyen, High School for Gifted Students, HUS, Vietnam Contact If you have any question about this ebook, please contact us. Email: nguyendunghus@gmail.com Acknowledgments We thank a lot to Mathlinks Forum 1 and their members for the reference to problems and many nice solutions from them! Hanoi, 10 October 2008 1 Website: http://mathlinks.ro Inequalities from 2008 Mathematical Competition Abbreviations • IMO International mathematical Olympiad • TST Team Selection Test • MO Mathematical Olympiad • LHS Left hand side • RHS Right hand side • W.L.O.G Without loss of generality • : cyclic Contents 1 Problems 4 2 Solutions 10 3 Chapter 1 Problems Pro 1. (Vietnamese National Olympiad 2008) Let x, y, z be distinct non-negative real numbers. Prove that 1 (x − y) 2 + 1 (y − z) 2 + 1 (z − x) 2 ≥ 4 xy + yz + zx . ∇ Pro 2. (Iranian National Olympiad (3rd Round) 2008). Find the smallest real K such that for each x, y, z ∈ R + : x √ y + y √ z + z √ x ≤ K (x + y)(y + z)(z + x) ∇ Pro 3. (Iranian National Olympiad (3rd Round) 2008). Let x, y, z ∈ R + and x + y + z = 3. Prove that: x 3 y 3 + 8 + y 3 z 3 + 8 + z 3 x 3 + 8 ≥ 1 9 + 2 27 (xy + xz + yz) ∇ Pro 4. (Iran TST 2008.) Let a, b, c > 0 and ab +ac +bc = 1. Prove that: √ a 3 + a + √ b 3 + b + √ c 3 + c ≥ 2 √ a + b + c ∇ 4 Inequalities from 2008 Mathematical Competition Pro 5. (Macedonian Mathematical Olympiad 2008.) Positive num- bers a, b, c are such that (a + b) (b + c) (c + a) = 8. Prove the inequality a + b + c 3 ≥ 27 a 3 + b 3 + c 3 3 ∇ Pro 6. (Mongolian TST 2008) Find the maximum number C such that for any nonnegative x, y, z the inequality x 3 + y 3 + z 3 + C(xy 2 + yz 2 + zx 2 ) ≥ (C + 1)(x 2 y + y 2 z + z 2 x). holds. ∇ Pro 7. (Federation of Bosnia, 1. Grades 2008.) For arbitrary reals x, y and z prove the following inequality: x 2 + y 2 + z 2 − xy − yz − zx ≥ max{ 3(x − y) 2 4 , 3(y − z) 2 4 , 3(y − z) 2 4 }. ∇ Pro 8. (Federation of Bosnia, 1. Grades 2008.) If a, b and c are positive reals such that a 2 + b 2 + c 2 = 1 prove the inequality: a 5 + b 5 ab(a + b) + b 5 + c 5 bc(b + c) + c 5 + a 5 ca(a + b) ≥ 3(ab + bc + ca) − 2 ∇ Pro 9. (Federation of Bosnia, 1. Grades 2008.) If a, b and c are positive reals prove inequality: (1 + 4a b + c )(1 + 4b a + c )(1 + 4c a + b ) > 25 ∇ Pro 10. (Croatian Team Selection Test 2008) Let x, y, z be positive numbers. Find the minimum value of: (a) x 2 + y 2 + z 2 xy + yz (b) x 2 + y 2 + 2z 2 xy + yz Inequalities from 2008 Mathematical Competition ∇ Pro 11. (Moldova 2008 IMO-BMO Second TST Problem 2) Let a 1 , . . . , a n be positive reals so that a 1 + a 2 + . . . + a n ≤ n 2 . Find the minimal value of A = a 2 1 + 1 a 2 2 + a 2 2 + 1 a 2 3 + . . . + a 2 n + 1 a 2 1 ∇ Pro 12. (RMO 2008, Grade 8, Problem 3) Let a, b ∈ [0, 1]. Prove that 1 1 + a + b ≤ 1 − a + b 2 + ab 3 . ∇ Pro 13. (Romanian TST 2 2008, Problem 1) Let n ≥ 3 be an odd integer. Determine the maximum value of |x 1 − x 2 | + |x 2 − x 3 | + . . . + |x n−1 − x n | + |x n − x 1 |, where x i are positive real numbers from the interval [0, 1] ∇ Pro 14. (Romania Junior TST Day 3 Problem 2 2008) Let a, b, c be positive reals with ab + bc + ca = 3. Prove that: 1 1 + a 2 (b + c) + 1 1 + b 2 (a + c) + 1 1 + c 2 (b + a) ≤ 1 abc . ∇ Pro 15. (Romanian Junior TST Day 4 Problem 4 2008) Determine the maximum possible real value of the number k, such that (a + b + c) 1 a + b + 1 c + b + 1 a + c − k ≥ k for all real numbers a, b, c ≥ 0 with a + b + c = ab + bc + ca. ∇ Inequalities from 2008 Mathematical Competition Pro 16. (Serbian National Olympiad 2008) Let a, b, c be positive real numbers such that x + y + z = 1. Prove inequality: 1 yz + x + 1 x + 1 xz + y + 1 y + 1 xy + z + 1 z ≤ 27 31 . ∇ Pro 17. (Canadian Mathematical Olympiad 2008) Let a, b, c be positive real numbers for which a + b + c = 1. Prove that a − bc a + bc + b − ca b + ca + c − ab c + ab ≤ 3 2 . ∇ Pro 18. (German DEMO 2008) Find the smallest constant C such that for all real x, y 1 + (x + y) 2 ≤ C · (1 + x 2 ) · (1 + y 2 ) holds. ∇ Pro 19. (Irish Mathematical Olympiad 2008) For positive real num- bers a, b, c and d such that a 2 + b 2 + c 2 + d 2 = 1 prove that a 2 b 2 cd + +ab 2 c 2 d + abc 2 d 2 + a 2 bcd 2 + a 2 bc 2 d + ab 2 cd 2 ≤ 3/32, and determine the cases of equality. ∇ Pro 20. (Greek national mathematical olympiad 2008, P1) For the positive integers a 1 , a 2 , , a n prove that n i=1 a 2 i n i=1 a i kn t ≥ n i=1 a i where k = max {a 1 , a 2 , , a n } and t = min {a 1 , a 2 , , a n }. When does the equality hold? ∇ Inequalities from 2008 Mathematical Competition Pro 21. (Greek national mathematical olympiad 2008, P2) If x, y, z are positive real numbers with x, y, z < 2 and x 2 + y 2 + z 2 = 3 prove that 3 2 < 1 + y 2 x + 2 + 1 + z 2 y + 2 + 1 + x 2 z + 2 < 3 ∇ Pro 22. (Moldova National Olympiad 2008) Positive real numbers a, b, c satisfy inequality a + b + c ≤ 3 2 . Find the smallest possible value for: S = abc + 1 abc ∇ Pro 23. (British MO 2008) Find the minimum of x 2 + y 2 + z 2 where x, y, z ∈ R and satisfy x 3 + y 3 + z 3 − 3xyz = 1 ∇ Pro 24. (Zhautykov Olympiad, Kazakhstan 2008, Question 6) Let a, b, c be positive integers for which abc = 1. Prove that 1 b(a + b) ≥ 3 2 . ∇ Pro 25. (Ukraine National Olympiad 2008, P1) Let x, y and z are non-negative numbers such that x 2 + y 2 + z 2 = 3. Prove that: x x 2 + y + z + y x + y 2 + z + z x + y + z 2 ≤ √ 3 ∇ Pro 26. (Ukraine National Olympiad 2008, P2) For positive a, b, c, d prove that (a + b)(b + c)(c + d)(d + a)(1 + 4 √ abcd) 4 ≥ 16abcd(1 + a)(1 + b)(1 + c)(1 + d) ∇ Inequalities from 2008 Mathematical Competition Pro 27. (Polish MO 2008, Pro 5) Show that for all nonnegative real values an inequality occurs: 4( √ a 3 b 3 + √ b 3 c 3 + √ c 3 a 3 ) ≤ 4c 3 + (a + b) 3 . ∇ Pro 28. (Chinese TST 2008 P5) For two given positive integers m, n > 1, let a ij (i = 1, 2, ··· , n, j = 1, 2, ··· , m) be nonnegative real numbers, not all zero, find the maximum and the minimum values of f, where f = n n i=1 ( m j=1 a ij ) 2 + m m j=1 ( n i=1 a ij ) 2 ( n i=1 m j=1 a ij ) 2 + mn n i=1 m i=j a 2 ij ∇ Pro 29. (Chinese TST 2008 P6) Find the maximal constant M, such that for arbitrary integer n ≥ 3, there exist two sequences of positive real number a 1 , a 2 , ··· , a n , and b 1 , b 2 , ··· , b n , satisfying (1): n k=1 b k = 1, 2b k ≥ b k−1 + b k+1 , k = 2, 3, ··· , n −1; (2):a 2 k ≤ 1 + k i=1 a i b i , k = 1, 2, 3, ··· , n, a n ≡ M. Chapter 2 Solutions Problem 1. (Vietnamese National Olympiad 2008) Let x, y, z be dis- tinct non-negative real numbers. Prove that 1 (x − y) 2 + 1 (y − z) 2 + 1 (z − x) 2 ≥ 4 xy + yz + zx . Proof. (Posted by Vo Thanh Van). Assuming z = min{x, y, z}. We have (x − z) 2 + (y − z) 2 = (x − y) 2 + 2(x − z)(y − z) So by the AM-GM inequality, we get 1 (x − y) 2 + 1 (y − z) 2 + 1 (z − x) 2 = 1 (x − y) 2 + (x − y) 2 (y − z) 2 (z − x) 2 + 2 (x − z)(y − z) ≥ 2 (x − z)(y − z) + 2 (x − z)(y − z) = 4 (x − z)(y − z) ≥ 4 xy + yz + zx We are done. Proof. (Posted by Altheman). Let f(x, y, z) denote the LHS minus the RHS. Then f(x + d, y + d, z + d) is increasing in d so we can set the least of x + d, y + d, z + d equal to zero (WLOG z = 0). Then we have 1 (x − y) 2 + 1 x 2 + 1 y 2 − 4 xy = (x 2 + y 2 − 3xy) 2 x 2 y 2 (x − y) 2 ≥ 0 ∇ 10 [...]... the inequality It becomes x3 + y 3 + cx2 y ≥ (c + 1)xy 2 Thus, we have to find the minimal value of f (y) = 1 y3 − y2 + 1 1 =y+ 2−y y y(y − 1) You can see here: http://www.mathlinks.ro/viewtopic.php?p =113 0901 Inequalities from 2008 Mathematical Competition when y > 1 It is easy to find that f (y) = 0 ⇔ 2y − 1 = (y(y − 1))2 ⇔ y 4 − 2y 3 + y 2 − 2y + 1 = 0 Solving this symmetric equation gives us: √ 1... value is 8/3 Expanding x− we get x2 + y 2 + 2z 2 − 2 y 3 2 8/3xy − + √ 1 y − 6z 3 2 ≥ 0, 8/3yz ≥ 0, so x2 + y 2 + z 2 ≥ xy + yz Equality occurs, for example, if x = 2, y = 8 3 √ 6, and z = 1 Problem 11 (Moldova 2008 IMO-BMO Second TST Problem 2) Let a1 , , an be positive reals so that a1 +a2 + .+an ≤ n Find the minimal 2 value of 1 1 1 A = a2 + 2 + a2 + 2 + + a2 + 2 1 2 n a2 a3 a1 Proof (Posted... where n f= ( n i=1 ( n i=1 m 2 j=1 aij ) m 2 j=1 aij ) +m + mn m j=1 ( n i=1 n 2 i=1 aij ) m 2 i=j aij Proof (Posted by tanpham) We will prove that the maximum value of f is 1 • For n = m = 2 Setting a11 = a, a21 = b, a12 = x, a21 = y We have 2 ((a + b)2 + (x + y)2 + (a + x)2 + (b + y)2 ) f= ≤1 (a + b + x + y)2 + 4 (a2 + b2 + x2 + y 2 ) ⇔= (x + b − a − y)2 ≥ 0 as needed • For n = 2, m = 3 Using the . yz (b) x 2 + y 2 + 2z 2 xy + yz Inequalities from 2008 Mathematical Competition ∇ Pro 11. (Moldova 2008 IMO-BMO Second TST Problem 2) Let a 1 , . . . , a n be positive reals so that. = y 3 − y 2 + 1 y 2 − y = y + 1 y(y − 1) 1 You can see here: http://www.mathlinks.ro/viewtopic.php?p =113 0901 Inequalities from 2008 Mathematical Competition when y > 1. It is easy to find. y 2 + z 2 xy + yz ≥ 8 3 . Equality occurs, for example, if x = 2, y = √ 6, and z = 1. ∇ Problem 11. (Moldova 2008 IMO-BMO Second TST Problem 2) Let a 1 , . . . , a n be positive reals so that